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In this lesson, we will talk about Transmission
Line Parameters, before I go into the
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transmission line parameters itself.
.
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We will first talk about the questions. That
I ask at the end of lesson 2. Well, the first
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question was, what is a single line diagram?
Well, the answer to that question is, single
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line diagrams are concise way of communicating
the basis arrangement of power system
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components. Single line diagrams use a single
line to represent all the three phases. And
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they show the relative electrical connections
of various electrical components, which are
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used in assembling the power system.
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..
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The second question was, how base values of
voltages are chosen on two sides of a
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transformer? Well, the answer of this question
is, ratio of base values on either side of
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a
transformer is selected to be the same as
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the ratio of the transformer voltage ratings.
Well, this is necessary, because when we use
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this voltage basis on the two sides as the
say as the voltage ratio of the transformer.
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Then, we find that the per unit impedance
on
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the two sides of the transformer are same.
And therefore, it eliminates the use of the
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ideal
transformer in the process steps. This helps
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considerably in reducing the calculations
for
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the power system circuit is…
.
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.Well, the third question was, for a 500 Mega
Watt 22 kV generator the per unit
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impedance is 0.8 per unit on it is own base.
What does it is per unit impedance at 100
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Mega Watt and 33 kV base? Well, as we had
seen in lessons 2, the Z per unit on the new
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base is given as set per unit and the old
base multiplied by old V base divided by new
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V
base square multiplied by new m V a base divided
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by old m V a base.
So, using this relationship, once we substitute
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these values, we will get set per unit at
the
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new base of 100 Mega Watt and 33 kV as pointed
into 22 by 33 whole square multiplied
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by 100 by 500, which will come out to be equal
to 0.0711 per unit. That is from 0.8 per
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unit, the values changes to 0.0711 per unit
on the new base.
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.
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So, after this, we will get into the lessons
3. In this, we will be talking about the
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transmission line parameters. Well, what we
will do is, we will starts with introduction
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to
transmission line. Then, we will talk about
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the types of overhead conductors. We will
talk about, how to calculate resistance of
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the transmission line. And we will talk about
the calculation of inductance, for a single
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phase transmission line.
So, first we will go into the overhead transmission
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system. The overhead transmission
system is mostly used at high voltage. Mainly,
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because it is much cheaper, compared to
an underground system. Now, the type of conductance,
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which we use in overhead
transmission are based on the kind of conducting
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material. That we are using as well as
the strength, which we need for these overhead
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conductors.
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..
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So, the conductors, which are used, are generally
copper conductors. Nowadays, copper
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conductors are hardly chosen or they are almost
non-extent. They are basically extent;
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we do not use them at all. Mostly, because
copper is very expensive and it supply is
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quite
limited. Well, the type of conductors, which
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are used mostly, is called ACSR conductors.
That is Aluminum Conductor Steel Reinforced.
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This is mainly because; aluminum is
much cheaper as well as it is available in
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abundance.
Aluminum has much lower conductivity as compared
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to copper, which means, that we
have to use larger cross sectional area, for
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the same current carrying capacity of the
line
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or the conductor. Another problem with aluminum
conductor is, that it is tensile strength
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is not very high. And therefore, it needs
to be reinforce, that is it has to have larger
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tensile strength to take care of it is own
weight or to make it more robust. When, we
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string it between the tables.
This is provided by steel reinforcement, which
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is done by putting steel strands at the core
and the aluminum strands of conductor is put
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on top of this steel core. Other types of
conductors, which are not so much in use,
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are All Aluminum-Alloy Conductor. Here
again, aluminum alloy is used, because it
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gives you more strength to weight ratio. That
is, it provides higher strength, for the same
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weight as compared to all aluminum
conductors. So, this is also used some time,
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but it is somewhat more expensive. Other
types are ACAR, that is Aluminum Conductor
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Alloy reinforced.
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.Now, this is again the same thing instead
of steel reinforcement. We have aluminum
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alloy reinforcement provided. One type of
conductor, which is used for extra high
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voltage lands is called expanded ACSR conductor.
This is same as the aluminum
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conductor is steel reinforce, except that,
between the steel core and the aluminum
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conductors. We provide some filler material,
such as paper or fiber; this increases the
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effective diameter of the conductor, for same
conducting cross sectional area.
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This has an advantage that with the larger
diameter, the electrical stress, for these
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high
voltage conductors, extra high voltage conductors
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is reduced. And their by, it reduces
corona losses. Also, this increase in the
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diameter reduces the inductance of the conductor
also to some extent. We will talk about these
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later. Well, the conductors let we, generally
use are always stranded conductors.
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.
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Here, I am showing you a diagram or a cross
section for an ACSR conductor, where we
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are using the steel reinforcement or the steel
strands at the core or a center for the
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conductor. And the aluminum strands are put
on top of this. Now, why do, we use
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stranded conductor. One of the reasons for
this is, that manufacturing or making these
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stranded conductors is much easier, especially
if the size is larger. Because, for larger
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sizes, what you need is keep on adding layers
of strand. Therefore, manufacturing
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becomes much easier to build.
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.The other advantage is, that these stranded
conductors, provide better mechanical
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strength as well as better handling. That
is, they are much more flexible as compared
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to a
solid conductor of the same diameter or the
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same cross sectional area. Therefore, they
are prepared over single solid conductor.
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.
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In this diagram, we have taken the view of
this stranded conductor. Again, here you find
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that the steel strands are at the center and
we have layers of strands of conductor,
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aluminum conductor on top of it. Now, aluminum
has much higher conductivity as
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compared to steel. And therefore, most of
the current flows through the aluminum
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conductor and hardly any current flows through
the steel strands.
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Here, one more thing, you can notice, that
is the strands or the layer of, there is
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conductors strands are spiral in opposite
direction. This is only to provide a better
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binding for these conductors. That is the
strands, do not open up easily. So, this provides
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a better binding and better mechanical strand.
That is why, spiraling is there. Next, we
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talk about the support structure, which is
normally transmission tower or poles.
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..
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Now, here you can see, that we have we use
different kind of structure depending on the
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voltage level. So, at lower voltage level,
we can use this kind of a structure. Also,
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we see
that for this, we are using a single conductor.
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Whereas, in this structure, we are seeing,
we are using two conductors for each phase.
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This kind of an arrangement is called
bundle conductors. This is generally used
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at high voltages of 220 kV and above.
The reason behind this is, this increases
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the effective radius of the conductors. Therefore,
it reduces the electrical stress and so the
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corona losses. And this also reduces the
inductance, which is very much useful in terms
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of reducing the voltage drop. Now, in
this structure, we also find that at top of
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the structures are the towers or the poles,
we use
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ground wires.
Now, these ground wires are provided mainly
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to protect the phase was from the direct
lightening strike. What happens, when since,
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these are at the top, the lightening will
normally strike these ground wires. And these
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wires are grounded at the tower potting
with low resistance. Therefore, when a lightning
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strikes, it harmlessly flows to the
ground without effecting or damaging the phase
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conductors at all.
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..
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Now, conductors can be placed in various arrangements.
The earlier diagram was
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showing the arrangement of conductors in a
horizontal plane. Many times, we arrange
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the conductors in different fashion, like
in vertical plane or in a hexagonal plane.
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Here,
this tower is showing a double circuit line.
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Where, the three phase conductor a, b, c of
one circuit is on this side and a, b, c of
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the other circuit is on the other side.
This kind of arrangement is also used for
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especially putting this, a, b, c here. And
putting
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a, b, c in this fashion, this again reduces
the inductance of the system. Sometimes, when
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the load on the system has not grown, what
we use is, we use a single circuit line. But,
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with a tower, which can be expanded to take
care of double circuit line, when the load
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builds up. Now, when we are stranding these
conductors on towers, we have to take care
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of certain mechanical considerations also.
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..
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Whenever, we strand these conductors, we find
that, there is some sag, which occurs at
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the middle of the span of the towers. Now,
this sag depends on the tension. That these
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conductors can take. And therefore, we cannot
provide very large amount of tension to
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reduce this sag. Another aspect of the sag
is, that this governs, what is the minimum
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ground clearance. That is available.
Now, this ground clearance has to be there
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to avoid the voltage stress for to any person
or any object, which is on the ground. It
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also has to be there, if people are any vehicle
is
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moving, if that is also will get stress by
this electrical field. And therefore, a certain
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minimum clearance is required. This is given
by certain standards, like for a forward kV
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line. You need around 12 meters of ground
clearance. That is the minimum ground
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clearance should be around 12 meters.
This clearance again governs, what is going
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to be the length between the two towers or
the span of the conductor. Another mechanical
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aspect that we need to consider is
vibration. Because, when flowing across the
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conductors, there will be vibrations. And
therefore, we that introduces fatigue in the
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conductors as well as tower. And they
become brittle and can get damage.
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Therefore, we need to damp these vibrations
and for this, it is also done by adjusting
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the
tension. That is one way of doing it and the
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other is putting vibration dampers in the
on
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the conductors, which absorbs the energy.
Also, when these transmission conductors are
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.subjected to across air flow, then they vibrate.
And because of this vibration, one has to
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see, that the minimum distance between the
two phase conductors is always maintained.
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And this governs, the spacing between the
conductors, like for a 400 kV line. The
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conduct spacing between two phase conductors
must be of the order 10 meters or so.
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.
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Let us we will talk about the electrical parameters.
Now, we are using conductor, over a
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conductors and these conductors has certain
amount of resistance to the current flowing
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through them. So, depending on this resistance,
there is going to be power loss in these
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conductors. Therefore, we must find out, what
is the resistance of the conductor. Then,
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we will also talk about inductance.
This comes because, once the current flows
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in the conductor, it sets of magnetic field.
Because, of which there is certain amount
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of inductance. Since, these conductors are
at
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high voltage. There is voltage difference
between the two phase conductors as well as
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between phase conductors and ground. Therefore,
there is capacitance involved between
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them. So, we will also talk about capacitance.
So, these three parameters are the most
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important parameters for a transmission line
electrical modeling.
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..
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Well, as per as resistances concerned, the
dc resistance of a conductor at any temperature
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is given by R d c is equal to rho l by A.
Where, rho is a function of time, sorry, rho
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is a
function of temperature. And that is the resistivity
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of the conductor changes with
temperature. If the temperature goes up, resistivity
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will also increase. L is the length of
the conductor and A is the conductor cross
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sectional area. If we take length in meter
and
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cross sectional are in meter square. And then
we get this resistance in ohms. Where, rho
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T is given in ohm meter.
.
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.Now, resistivity of a conductor material
varies linearly over normal operating
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temperature, according to the relation. Rho
at temperature T 2 is equal to rho at
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temperature T 1 into T 2 plus T divided by
T 1 plus T. Where, this T is a constant or
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a
temperature constant. And this value is different
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for different material as we will see in
the table shown.
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This is about the d c resistance. But, the
ac resistance is somewhat higher than the
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dc
resistance. This is mainly, because at higher
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frequency, the current density in the
conductor is no longer uniform. In fact, higher
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the frequency less and less current will be
flowing through the center of the conductor.
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And more will be flowing through the outer
part of the conductor; this is termed as skin
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effect. And this reduces the effective
conductor cross sectional area. And therefore,
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the resistance of conductor and a
alternating current is somewhat higher.
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Normally, we find out the ac resistance from
the relationship. That the power loss
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divided by I square, gives us the resistance,
ac resistance of the conductor. That is, you
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can measure the power loss and you can measure
the current and from that, we can find
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out the ac resistance of the conductor. In
fact, most of the manufactures, who
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manufacture these conductors, who they manufacture
them in standard sizes. And there
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are table available, which provide the ac
resistance of the conductor per kilo meter
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length.
.
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.As I said, this table shows the temperature
constant values. It is, if we can see from
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this
table, which provides also the conductivity.
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The copper, annealed copper has the input
as
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100 percent conductivity. Then, the hard-drawn
aluminum, which is used for a conductor
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has 61 percent conductivity as compared to
annealed copper. For finding out, the
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resistance, dc resistance of the conductor
at different temperatures, we use this
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temperature constant T. Now, conductor resistance
depends on certain factor as we have
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seen. It depends on temperature, because the
resistivity increases, when the temperature
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increases.
.
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It also depends on spiraling, because we are
using standard conductors. And we are seen,
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that these conductors has spiral in opposite
direction to provide a better binding from
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the
strands. And therefore, this spiraling effectively
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increases the actual length of the
conductors. That is, the length of the conductor
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and actual length of the strand is not
same. The length of strand will be one or
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two percent more than that of the conductors
as
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such.
And therefore, we have to take care of this
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larger length and that increases the resistance.
That is resistance, due to spiraling will
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also increase in the same order of one or
two
209
00:25:05,450 --> 00:25:14,509
person. We have also said that, because of
high frequency, the current density does not
210
00:25:14,509 --> 00:25:20,869
remain same, because more and more current
flows, through the outer part rather than
211
00:25:20,869 --> 00:25:29,860
.the core. And therefore, the effective cross
sectional area is reduced and this also
212
00:25:29,860 --> 00:25:34,480
increases the resistance of the conductor.
.
213
00:25:34,480 --> 00:25:44,260
Next, we will talk about the inductance. Well,
for calculating inductance, we need to go
214
00:25:44,260 --> 00:25:52,379
through four steps. First is, to find out
the magnetic field intensity H using Ampere’s
215
00:25:52,379 --> 00:26:01,889
law. Second step is, from H, we can find out
the flux density, B is equal to mu H. Now,
216
00:26:01,889 --> 00:26:09,990
since, we are using aluminum or copper conductors
mostly, mu of this is going to be
217
00:26:09,990 --> 00:26:17,629
equal to mu 0, because these conductors are
non magnetic. So, permeability for these
218
00:26:17,629 --> 00:26:25,350
conductors is same as that of air, which is
equal to 4 pi into 10 to power minus 7.
219
00:26:25,350 --> 00:26:30,960
Now, flux linkages once we have calculated
flux density, we can calculated the flux
220
00:26:30,960 --> 00:26:39,360
linkages for the current, which is flowing
in that conductor. And finally, we can find
221
00:26:39,360 --> 00:26:44,070
out
inductance from the flux linkages as inductance
222
00:26:44,070 --> 00:26:49,499
is equal to flux linkage per unit current.
223
00:26:49,499 --> 00:26:50,499
..
224
00:26:50,499 --> 00:27:02,809
So, let us take first a solid conductor. Here,
we have taken one conductor, which is of
225
00:27:02,809 --> 00:27:13,880
cross sectional area with a radius r and 1
meter in length. Now, in this conductor is
226
00:27:13,880 --> 00:27:23,279
carrying current I. Since, the current cannot
flow in just one conductor, there has to be
227
00:27:23,279 --> 00:27:25,049
a
returned conductor. Here, we are assuming
228
00:27:25,049 --> 00:27:32,119
the return conductor is at a very far away
distance. So, it does not affect the magnetic
229
00:27:32,119 --> 00:27:36,480
field produce by this conductor. That is at
a
230
00:27:36,480 --> 00:27:41,869
infinite distance from this conductor.
So, that is one assumption that we are making.
231
00:27:41,869 --> 00:27:49,309
Another assumption that we make here in
finding inductance is that the current density
232
00:27:49,309 --> 00:27:58,350
inside the conductor is uniform. Though,
we have seen that, the current density at
233
00:27:58,350 --> 00:28:03,399
higher frequencies is not uniform. But, at
lower
234
00:28:03,399 --> 00:28:10,860
frequency is a 50 hertz or 60 hertz, it is
more or less uniform. So, this assumption
235
00:28:10,860 --> 00:28:13,630
is
generally made.
236
00:28:13,630 --> 00:28:21,389
Now, we can see that, for this conductor,
since the current is flowing inside the
237
00:28:21,389 --> 00:28:32,610
conductor. The current will produce magnetic
field inside the conductor also. And
238
00:28:32,610 --> 00:28:37,289
because of the current flowing, there will
be magnetic field outside the conductor also.
239
00:28:37,289 --> 00:28:43,399
Therefore, for calculating the inductance,
we will have to find out the flux linkages,
240
00:28:43,399 --> 00:28:48,139
for
the magnetic field, which is inside the conductor.
241
00:28:48,139 --> 00:28:54,019
And flux linkages for the magnetic
field, outside the conductor. That is, we
242
00:28:54,019 --> 00:28:58,380
need to do two parts and add them up to get
the
243
00:28:58,380 --> 00:29:00,330
final flux linkage.
244
00:29:00,330 --> 00:29:08,009
.So, first we will take the internal flux
linkages. Now, here, what we have done is,
245
00:29:08,009 --> 00:29:15,350
we
have taken circular contour at a distance
246
00:29:15,350 --> 00:29:27,120
x, from the center. And let H x be the intensity
of the magnetic field at any point on this
247
00:29:27,120 --> 00:29:31,049
contour. Since, this H x will be, this contour
is
248
00:29:31,049 --> 00:29:38,429
concentric, it is uniform with respect to
this conductor. This H x will also be same,
249
00:29:38,429 --> 00:29:46,080
everywhere on this contour and it is direction
will be tangential to this contour.
250
00:29:46,080 --> 00:29:47,080
.
251
00:29:47,080 --> 00:29:57,690
Therefore, from Ampere’s law, we know that
tangential component of H into d l integral,
252
00:29:57,690 --> 00:30:05,559
over a close contour is equal to the current
in closed. So, using this relationship for
253
00:30:05,559 --> 00:30:09,559
this
contour, we get H x into twice pi x, which
254
00:30:09,559 --> 00:30:13,870
is the contour length, when we integrated.
So,
255
00:30:13,870 --> 00:30:21,679
H x, x the tangential component of H is H
x and integral dl over the contour is twice
256
00:30:21,679 --> 00:30:30,710
pi x.
This is equal to I x, for x less than r. From
257
00:30:30,710 --> 00:30:33,549
this relationship, we can calculate H x is
equal
258
00:30:33,549 --> 00:30:50,580
to I x by twice pi x, ampere per meter.
Now, this H x field intensity H x is enclosing
259
00:30:50,580 --> 00:30:57,749
current I x. And what is this current I x?
This current I x, since we are assumed uniform
260
00:30:57,749 --> 00:31:04,379
current density throughout the conductor
is x by r whole square into I fraction of
261
00:31:04,379 --> 00:31:05,379
the total current I.
262
00:31:05,379 --> 00:31:06,379
..
263
00:31:06,379 --> 00:31:18,620
Therefore, we can write H x is equal to x
into I by twice pi r square. Because, for
264
00:31:18,620 --> 00:31:30,929
substituting for I x, x by r whole square
into I and in this position, we will get H
265
00:31:30,929 --> 00:31:33,960
x is
equal to x I by twice pi r square ampere per
266
00:31:33,960 --> 00:31:39,820
meter. And from this field intensity, we can
find out the flux density B x, which will
267
00:31:39,820 --> 00:31:47,090
be equal to mu 0 H x, as we are using non
magnetic conducting material. So, mu 0 H x,
268
00:31:47,090 --> 00:31:50,690
this is equal to mu 0 x into I by twice pi
r
269
00:31:50,690 --> 00:31:53,659
square Webber per meter square.
.
270
00:31:53,659 --> 00:32:08,679
.Now, if we look at this strip, which is 1
meter long and of thickness d x, at this point
271
00:32:08,679 --> 00:32:15,549
x
from the center. How much flux will be passing
272
00:32:15,549 --> 00:32:23,659
through this strip? Since, the flux lines
are concentric. And we know the flux density
273
00:32:23,659 --> 00:32:33,119
the area, which will be involved for the
flux lines will be B x into 1. Therefore,
274
00:32:33,119 --> 00:32:40,779
d phi is equal to mu 0 x into I, by twice
pi r
275
00:32:40,779 --> 00:32:51,080
square into d x into 1, which is the area.
And this flux is linking how much current?
276
00:32:51,080 --> 00:32:54,080
This flux is linking only the current which
is
277
00:32:54,080 --> 00:33:02,570
enclose by x. And that current is equal to
how much? That is pi x square by pi r square
278
00:33:02,570 --> 00:33:10,630
fraction of the total current Therefore, we
have the flux linkage is equal to pi x square
279
00:33:10,630 --> 00:33:14,450
by
pi r square into d phi. This is equal to mu
280
00:33:14,450 --> 00:33:20,489
0 I by twice pi r 4 x cubes d x Weber therms
per meter.
281
00:33:20,489 --> 00:33:21,489
.
282
00:33:21,489 --> 00:33:29,859
Now, if we integrate from the center to the
tip of the conductor. That is from 0 to r,
283
00:33:29,859 --> 00:33:32,989
this
flux linkage in this strip, we will get the
284
00:33:32,989 --> 00:33:37,460
total internal flux linkage of the conductor
for
285
00:33:37,460 --> 00:33:46,109
current flowing inside it. And this is equal
to mu 0 I by twice pi r 4 integral from 0
286
00:33:46,109 --> 00:33:51,409
to r of
x cube d x, which finally comes out to be
287
00:33:51,409 --> 00:33:56,399
half into 10 to power minus 7 I Weber therms
per meter.
288
00:33:56,399 --> 00:34:06,309
So, the internal flux linkage is constant
that is half into 10 to power minus 7 into
289
00:34:06,309 --> 00:34:13,470
I. So,
depends on the current, which is flowing and
290
00:34:13,470 --> 00:34:20,070
it has nothing to do with the radius of the
conductor. Now, next we have to find out the
291
00:34:20,070 --> 00:34:22,960
flux linkages outside the conductor.
292
00:34:22,960 --> 00:34:23,960
..
293
00:34:23,960 --> 00:34:30,250
Now, again for the same conductor, we have
taken a point, which is at a distance x from
294
00:34:30,250 --> 00:34:38,139
the center of the conductor, where x is greater
than r. Now, again we can find out take a
295
00:34:38,139 --> 00:34:46,679
concentrate contour and using Ampere’s law,
we will get H x into twice pi x is equal to
296
00:34:46,679 --> 00:34:49,810
I.
Because, now, this contour is enclosing what
297
00:34:49,810 --> 00:34:54,390
current, all the total current, which is
flowing in this conductor? Therefore, H x
298
00:34:54,390 --> 00:34:57,090
into twice pi x is equal to I, from there
we will
299
00:34:57,090 --> 00:35:03,470
get H x is equal to I by twice x ampere per
meter for x greater than r.
300
00:35:03,470 --> 00:35:04,470
.
301
00:35:04,470 --> 00:35:13,430
.Again, we can find out the flux density at
that point, this will be equal to mu 0 H x,
302
00:35:13,430 --> 00:35:20,860
which is equal to 4 pi into 10 power minus
7 into I by twice pi x. This will be equal
303
00:35:20,860 --> 00:35:24,690
to 2
into 10 to power minus 7 I by x Weber per
304
00:35:24,690 --> 00:35:35,100
meter square. Now, once, again we take a
small strip of a 1 meters length and d x thickness.
305
00:35:35,100 --> 00:35:43,190
Then, we can find out the flux, through
this strip and this will be equal to B x into
306
00:35:43,190 --> 00:35:46,910
d x into 1. So, again substituting the value,
this
307
00:35:46,910 --> 00:35:55,040
comes out to be 2 into 10 to power minus 7
I by x into d x Weber per meter.
308
00:35:55,040 --> 00:35:56,040
.
309
00:35:56,040 --> 00:36:03,870
Now, if we want we can find out the total
flux linkages between any two points D 1 and
310
00:36:03,870 --> 00:36:10,720
D 2. From the center of the conductor, which
we can write as lambda 1 to is equal to
311
00:36:10,720 --> 00:36:18,260
integral D 1 to D 2 of d lambda, which will
be equal to 2 into 10 to power minus 7 I
312
00:36:18,260 --> 00:36:27,600
integral D 1 to D 2 d x by x. And which after
integrating comes out to be 2 into 10 to
313
00:36:27,600 --> 00:36:33,960
power minus 7, I log n D 2 by D 1 Weber therms
per meter.
314
00:36:33,960 --> 00:36:41,040
So, if we want to find out the flux linkage,
total flux linkage up to any point P, for
315
00:36:41,040 --> 00:36:45,190
this
conductor carrying current I. Then, we need
316
00:36:45,190 --> 00:36:52,360
to add the flux linkages internal to the
conductor and the flux linkage out to outside
317
00:36:52,360 --> 00:36:59,950
of the conductor. From the conductor
surface up to a distance D of this point p,
318
00:36:59,950 --> 00:37:04,930
from the center of the conductor.
This will be equal to half into 10 to power
319
00:37:04,930 --> 00:37:10,900
minus 7 I, which is the internal flux linkage.
Plus 2 into 10 to power minus 7 I log n D
320
00:37:10,900 --> 00:37:14,130
by r, which is the external flux linkage up
to a
321
00:37:14,130 --> 00:37:16,910
distance D from the conductor.
322
00:37:16,910 --> 00:37:17,910
..
323
00:37:17,910 --> 00:37:27,910
Now, using half is equal to 2 log n e to the
power 1 by 4. Substituting this, we get the
324
00:37:27,910 --> 00:37:35,700
flux linkage lambda p is equal to 2 into 10
to power minus 7 I log n e to the power 1
325
00:37:35,700 --> 00:37:43,060
by 4
plus log n D by r, which can be written as
326
00:37:43,060 --> 00:37:47,390
2 into 10 to power minus 7 I log n D by e
to
327
00:37:47,390 --> 00:37:56,530
the power minus 1 by 4 r. This we can write
as 2 into 10 to power minus 7 I log n D by
328
00:37:56,530 --> 00:38:05,150
dash. Where, r dash is e to the power minus
1 by 4 into r, which is equal to 0.7788 r.
329
00:38:05,150 --> 00:38:12,160
That is, what we are seeing is, because of
the internal flux linkages, the effective
330
00:38:12,160 --> 00:38:18,660
radius
of the conductor is now r dash or which is
331
00:38:18,660 --> 00:38:24,460
0.7788 r. That is the effective radius has
got
332
00:38:24,460 --> 00:38:34,190
somewhat produced as per as the flux linkages
are concerned. That is, if we are not
333
00:38:34,190 --> 00:38:43,380
considering any internal flux linkage, then
we have to use the relationship goes use as
334
00:38:43,380 --> 00:38:46,260
r
dash, which is the effective radius in that
335
00:38:46,260 --> 00:38:50,650
case.
So, therefore, once we have calculated the
336
00:38:50,650 --> 00:38:54,581
flux linkages up to that point P. Then, we
can
337
00:38:54,581 --> 00:39:00,410
also calculate the inductance, due to flux
linkage is up to that point P. That will be
338
00:39:00,410 --> 00:39:07,790
lambda P by I. And this is equal to 2 into
10 to power minus 7 log n D by r dash Henry’s
339
00:39:07,790 --> 00:39:11,160
per meter.
340
00:39:11,160 --> 00:39:12,160
..
341
00:39:12,160 --> 00:39:21,340
Well, we have now been able to find out the
inductance of a single conductor with a
342
00:39:21,340 --> 00:39:29,100
return, which is very far away from it. That
is not a normal practical case. Normally,
343
00:39:29,100 --> 00:39:35,420
we
will have conductors, which will be nearby
344
00:39:35,420 --> 00:39:44,560
only and so we can now take up a case,
where we have a system, where we have large
345
00:39:44,560 --> 00:39:49,610
number of conductors. Here we have
assumed n number of conductors is forming
346
00:39:49,610 --> 00:39:54,140
this system in which current is flowing.
.
347
00:39:54,140 --> 00:40:02,411
Now, since these conductors form the total
number of conductors in the system, one
348
00:40:02,411 --> 00:40:09,650
thing that we will see is that, sum of all
the currents in these conductor must be equal
349
00:40:09,650 --> 00:40:10,650
to
350
00:40:10,650 --> 00:40:16,070
.0. That is some conductors are carrying current
in one direction and some conductors are
351
00:40:16,070 --> 00:40:24,270
acting as the return conductors. So, this
is, what we will get, that is I 1 plus I 2
352
00:40:24,270 --> 00:40:30,530
plus up to
I N or sigma j is equal to 1 to N of I j will
353
00:40:30,530 --> 00:40:38,000
be equal to 0. That is sum of the all the
currents in the conductors is equal to 0.
354
00:40:38,000 --> 00:40:47,640
Now, using the earlier relationship, we can
find out the flux linkage for the current
355
00:40:47,640 --> 00:40:55,850
flowing in for the conductor k up to a point
P, due to current flowing in conductors k,
356
00:40:55,850 --> 00:41:09,030
which will be given by 2 into 10 to power
minus 7 I k log n D P k. That
357
00:41:09,030 --> 00:41:18,550
is the distance of
the conductor k from P divided by r dash,
358
00:41:18,550 --> 00:41:24,930
which is the radius of this conductor.
Similarly, if we want to find out, the flux
359
00:41:24,930 --> 00:41:31,910
linkage with this conductor, due to current
flows in some other conductor. We can write
360
00:41:31,910 --> 00:41:38,760
down this similar relationship, except that,
instead of this r dash, we will have to write
361
00:41:38,760 --> 00:41:45,140
D P j, sorry D k j.
.
362
00:41:45,140 --> 00:41:52,570
Therefore, if we use that, we can find out
the total flux linkage up to point P, for
363
00:41:52,570 --> 00:41:57,690
the
conductor k, will be equal to lambda k P 1.
364
00:41:57,690 --> 00:42:05,880
That is flux linkage of conductor k up to
point P, due to current flowing in conductor
365
00:42:05,880 --> 00:42:11,000
one and so on, for all the currents flowing
in
366
00:42:11,000 --> 00:42:17,020
different, all the N conductors. Therefore,
this is equal to 2 into 10 to power minus
367
00:42:17,020 --> 00:42:23,501
7
sigma j is equal to N I j log n D P j by D
368
00:42:23,501 --> 00:42:33,360
k j.
Now, here, we have used this relationship
369
00:42:33,360 --> 00:42:38,800
D k j and for the same conductor, that is
D k k,
370
00:42:38,800 --> 00:42:46,910
basically it is own distance will be equal
to r dash. Therefore, lambda flux linkage
371
00:42:46,910 --> 00:42:50,780
of
conductor k up to point P can be written as
372
00:42:50,780 --> 00:42:57,460
2 into 10 to power minus 7 sigma I j log n
1
373
00:42:57,460 --> 00:43:04,310
.by D k j plus 2 into 10 to power minus 7
sigma j is equal to 1 to N I j log n D P j.
374
00:43:04,310 --> 00:43:08,310
And
what we have done is, this term we have separated
375
00:43:08,310 --> 00:43:18,000
into two parts as two summations.
One is I j log n 1 by D k j plus I j log D
376
00:43:18,000 --> 00:43:19,000
P j.
.
377
00:43:19,000 --> 00:43:29,180
We will see why we do this. Therefore, lambda
k P is equal to 2 into 10 to power minus
378
00:43:29,180 --> 00:43:42,010
7 sigma I j log n 1 by D k j plus. Now, the
term that we had used here D P log n I j log
379
00:43:42,010 --> 00:43:46,280
n
D P j. Now, we are separating it into two
380
00:43:46,280 --> 00:43:51,900
parts for j is equal to 1 to N minus 1 I j
log n D
381
00:43:51,900 --> 00:44:00,990
p j plus for the N th conductor, we are separating
it a plus I N log n D P N. Now, we
382
00:44:00,990 --> 00:44:08,180
know, I N will be equal to minus of the sum
of N minus 1 conductor, because the sum of
383
00:44:08,180 --> 00:44:15,100
all the currents is equal to 0. That is I
N is equal to minus of sigma j is equal to
384
00:44:15,100 --> 00:44:17,110
1 to N
minus 1 I j.
385
00:44:17,110 --> 00:44:23,260
So, substituting this, here we get lambda
k P is equal to 2 into 10 to power minus 7
386
00:44:23,260 --> 00:44:30,060
sigma
j is equal to 1 to N I j log n, sorry, 1 by
387
00:44:30,060 --> 00:44:34,640
D k j. This plus should not be their plus
sigma j is
388
00:44:34,640 --> 00:44:45,780
equal to 1 to N minus 1 I j log n D P j minus
for this term sigma I j.
389
00:44:45,780 --> 00:44:46,780
..
390
00:44:46,780 --> 00:44:57,290
So, minus sigma I j j is equal to N minus
1 log n D P N, which can be written as 2 into
391
00:44:57,290 --> 00:45:07,690
10 to power minus 7 sigma I j j is equal to
1 to N log n 1 by D k j plus sigma I j log
392
00:45:07,690 --> 00:45:16,590
n D
P j by D P M, sorry, this should be N. This
393
00:45:16,590 --> 00:45:42,130
DPN term is coming. So, that will be DPN,
this is D P N. Now, if we take this point
394
00:45:42,130 --> 00:45:47,130
P very, very far away, then what happens is
this
395
00:45:47,130 --> 00:45:56,930
D P j and D P N will be almost same.
And therefore, this will D P j by D P M will
396
00:45:56,930 --> 00:46:06,020
become equal to 1 and the log n D P j by
DPM will be equal to 0. And therefore, finally,
397
00:46:06,020 --> 00:46:15,070
we will get the flux linkages, total flux
linkages for the conductor k equal to 2 into
398
00:46:15,070 --> 00:46:18,680
10 to power minus 7, sigma j is equal 1 to
N I
399
00:46:18,680 --> 00:46:26,710
j log n 1 by D k j. This is the final expression
that we get and this is a very important
400
00:46:26,710 --> 00:46:34,250
relationship for finding out the total flux
linkage for any conductor carrying current
401
00:46:34,250 --> 00:46:37,890
in a
system of conductors carrying current.
402
00:46:37,890 --> 00:46:38,890
...
403
00:46:38,890 --> 00:46:47,090
Now, we are try to use this, for a single
phase transmission line, which is consisting
404
00:46:47,090 --> 00:46:51,840
of
composite conductors. So, we have one conductor
405
00:46:51,840 --> 00:47:00,320
x and return conductor is y. Conductor
x is consisting of total N number for conductors
406
00:47:00,320 --> 00:47:07,490
each with a radius r x. That is same
radius conductors, but they are a total N
407
00:47:07,490 --> 00:47:14,070
number of conductors. And each conductor
carries 1 by N th of the total current. So,
408
00:47:14,070 --> 00:47:18,540
each conductor is the current is I, each
conductor is carrying I by N.
409
00:47:18,540 --> 00:47:27,280
Similarly, conductor y is consisting of total
M number of sub conductors and each of
410
00:47:27,280 --> 00:47:37,100
these conductors has sub conductors have a
radius r y. And they are carrying 1 by M part
411
00:47:37,100 --> 00:47:43,890
of the total current coming through. So, if
the current is flowing is I, then the return
412
00:47:43,890 --> 00:47:49,970
conductor will have currents I by M in each
of these sub conductors.
413
00:47:49,970 --> 00:47:50,970
..
414
00:47:50,970 --> 00:48:02,760
So, now we can find out the flux for any conductor
k in this system. In the conductor n is
415
00:48:02,760 --> 00:48:09,300
sub conductor k in the system of conductor
x. This will be given by 2 into 10 to power
416
00:48:09,300 --> 00:48:27,170
minus 7, I 1 I by N log n 1 by D k m minus
I by m sigma log n 1 by D k m. That is using
417
00:48:27,170 --> 00:48:34,400
this relationship for this set of conductors.
That is, what we are finding out for this
418
00:48:34,400 --> 00:48:40,500
conductor.
We are finding out the flux linkage, due to
419
00:48:40,500 --> 00:48:44,770
current in all the conductors in x as well
as all
420
00:48:44,770 --> 00:48:54,440
the conductor sub conductors in y. So, we
are adding all those fluxes. So, the total
421
00:48:54,440 --> 00:48:58,030
flux,
which is linking this conductor k will be
422
00:48:58,030 --> 00:49:02,290
given by I by N is the current flowing in
the all
423
00:49:02,290 --> 00:49:10,320
the sub conductors of conductor x. So, I by
N sigma log n 1 by D k m, where, D k m is
424
00:49:10,320 --> 00:49:16,580
the distance of all the conductors from the
conductor k.
425
00:49:16,580 --> 00:49:24,300
Similarly, the current minus I by m is flowing
in each sub conductor. So, the flux linkage
426
00:49:24,300 --> 00:49:31,590
is or the flux produce by them, which will
be linking this conductor will be given by
427
00:49:31,590 --> 00:49:36,430
I by
minus I by M sigma m is equal to 1 dash to
428
00:49:36,430 --> 00:49:47,900
M log n 1 by D k m. Now, what is the flux
linkage of this conductor? The current flowing
429
00:49:47,900 --> 00:49:56,000
in this conductor is I by N. So, total flux
linkage of this conductor will be one N th
430
00:49:56,000 --> 00:50:03,270
of the total flux. So, phi k by N is the flux
linkage of this conductor, this will be equal
431
00:50:03,270 --> 00:50:12,390
to 2 into 10 to power minus 7 into I. And
when we divide this expression by N we get
432
00:50:12,390 --> 00:50:21,960
1 by N square sigma log n 1 by D k m
minus one by N into M 1 by M again it gets
433
00:50:21,960 --> 00:50:26,201
divided by N. So, 1 by N into M sigma log
n
434
00:50:26,201 --> 00:50:30,070
1 by D k m.
435
00:50:30,070 --> 00:50:31,070
..
436
00:50:31,070 --> 00:50:39,850
Now, what will be the total flux linkage of
the conductor x, that will be the summation
437
00:50:39,850 --> 00:50:42,790
of
all the flux linkages of each sub conductor
438
00:50:42,790 --> 00:50:50,080
in that system of conductor x. Therefore,
lambda x is equal to summation K is equal
439
00:50:50,080 --> 00:50:53,530
to 1 to N lambda K, which will be equal to
2
440
00:50:53,530 --> 00:51:00,510
into 10 to power minus 7 I, sigma K is equal
to 1 to N of this terms. That is 1 by N
441
00:51:00,510 --> 00:51:08,840
square sigma log n 1 by D k m minus 1 by NM
sigma log n 1 by D k m. This is putting
442
00:51:08,840 --> 00:51:15,250
the expression lambda K here. And the summation
is from is equal to 1 to N, because we
443
00:51:15,250 --> 00:51:17,830
are talking for conductor x.
.
444
00:51:17,830 --> 00:51:28,980
.So, lambda x will be equal to 2 into 10 to
power minus 7 I log n. Now, here, what we
445
00:51:28,980 --> 00:51:42,680
have done is, summation of log n 1 by D k
m, can also be written as log n multiplication
446
00:51:42,680 --> 00:51:51,780
of 1 by D k m. So, we have used that relationship
and therefore, we have written 2 into
447
00:51:51,780 --> 00:51:58,320
10 to power minus 7 I, log n has been taken
out side. Therefore, it is multiplication
448
00:51:58,320 --> 00:52:04,230
K is
equal to 1 N into multiplication of m is equal
449
00:52:04,230 --> 00:52:11,810
to 1 dash to M D k m divided by
multiplication m is equal to 1 to N of D k
450
00:52:11,810 --> 00:52:15,790
m.
This will be to the power 1 by N square and
451
00:52:15,790 --> 00:52:25,400
this will be to the power 1 by N M, because
we had that 1 by N M and 1 by N square. Therefore,
452
00:52:25,400 --> 00:52:31,240
we can calculate L x, which will be
equal to 2 into 10 to power minus 7 log D
453
00:52:31,240 --> 00:52:42,420
x y by D x x. Where, this term we are writing
as D x y and this terms, we will we are writing
454
00:52:42,420 --> 00:52:43,470
as D x x.
.
455
00:52:43,470 --> 00:52:53,320
As we will show D x y is equal to M N th root
of sigma’s product k is equal to 1 to N
456
00:52:53,320 --> 00:53:02,380
and product m is equal to 1 dash to M D k
m. This total term D x y is called the G M
457
00:53:02,380 --> 00:53:08,590
D
or Geometric Mean Distance between the conductor
458
00:53:08,590 --> 00:53:18,550
x and y. Similarly, the term D x x,
that is the self GMD or Geometric Mean Radius
459
00:53:18,550 --> 00:53:26,130
is equal to the N square th root of
product k is equal to 1 to N into product
460
00:53:26,130 --> 00:53:33,810
m is equal to 1 to N of D k m.
That is product of each sub conductor’s
461
00:53:33,810 --> 00:53:38,990
distance to all the other distances. And this
is
462
00:53:38,990 --> 00:53:45,820
done for all the conductors, so you are getting
N square distances. And so we have the N
463
00:53:45,820 --> 00:53:50,330
square th root of this.
464
00:53:50,330 --> 00:53:51,330
..
465
00:53:51,330 --> 00:53:59,200
Similarly, we can calculate the inductance
of conductor system y, which will be equal
466
00:53:59,200 --> 00:54:03,670
to
2 into 10 to power minus 7 log n D x y by
467
00:54:03,670 --> 00:54:16,630
D y y Henry’s per meter per conductor. And
here, D y y will be equal to M square th root
468
00:54:16,630 --> 00:54:23,330
of product k is equal to 1 dash to M and
multiplied by product m is equal to 1 dash
469
00:54:23,330 --> 00:54:27,680
to M into D k m. That is the product of all
the
470
00:54:27,680 --> 00:54:35,210
distances from each sub conductor to other
sub conductor for all the sub conductors and
471
00:54:35,210 --> 00:54:44,320
conductor y. And the circuit inductance will
be the inductance of conductor x plus the
472
00:54:44,320 --> 00:54:50,150
inductance of conductor y. Because, that is
the total circuit, that we have consisting
473
00:54:50,150 --> 00:54:52,560
of
both the conductors in which the current is
474
00:54:52,560 --> 00:54:53,710
flowing.
.
475
00:54:53,710 --> 00:55:04,910
.So, once, we have calculated the inductance.
Now, we will close here and before, we
476
00:55:04,910 --> 00:55:12,730
close, I would you to answer the follow questions.
First is, what type of conductors is
477
00:55:12,730 --> 00:55:20,620
used for overhead transmission lines. Second
question is, how conductor diameter affects
478
00:55:20,620 --> 00:55:27,180
inductance of transmission line. And the third
question is, why bundled conductors are
479
00:55:27,180 --> 00:55:35,360
used in extra high voltage lines. So, I hope
after going through this lesson, you should
480
00:55:35,360 --> 00:55:37,710
be
able to answer these questions.
481
00:55:37,710 --> 00:55:38,710
Thank you.
Power System Analysis
482
00:55:38,710 --> 00:55:39,710
Prof. A.K.Sinha
Department of Electrical Engineering
483
00:55:39,710 --> 00:55:40,710
Indian Institute of Technology, Kharagpur
Lecture - 04
484
00:55:40,710 --> 00:55:57,170
Inductance Calculation
Welcome to lesson 4, on power system analysis.
485
00:55:57,170 --> 00:56:05,960
Before, we start this lesson 4, first I
would like to take up the questions, that
486
00:56:05,960 --> 00:56:08,410
we asked in lessons 3.
.
487
00:56:08,410 --> 00:56:16,260
.First question was, what are the types of
conductors used for overhead transmission
488
00:56:16,260 --> 00:56:21,400
line?
Well, some of the types of conductors used
489
00:56:21,400 --> 00:56:27,630
for overhead transmission lines are copper
conductors, which are very rarely used nowadays.
490
00:56:27,630 --> 00:56:35,030
ACSR, that is aluminium conductors
steel reinforced conductors. Then, all aluminum
491
00:56:35,030 --> 00:56:44,930
conductors are ACAR or expanded
ACSR conductors, which are used.
492
00:56:44,930 --> 00:56:54,080
As I have already said in lesson 3. Sometimes,
we want to increase the effective radius of
493
00:56:54,080 --> 00:57:01,860
the conductor and for that, we used expanded
ACSR conductor in EHV transmission
494
00:57:01,860 --> 00:57:11,230
lines. Now, Let us take an example for finding
out the inductance of a three phase
495
00:57:11,230 --> 00:57:12,230
transmission system.
.
496
00:57:12,230 --> 00:57:18,530
The example, that we are taking is for, the
conductor configuration of a completely
497
00:57:18,530 --> 00:57:25,500
transposed three phase overhead transmission
line, with bundle conductor is shown. All
498
00:57:25,500 --> 00:57:35,330
the conductors have a radius of 0.74 centimeter
with a 30 centimeter bundle space.
499
00:57:35,330 --> 00:57:36,330
..
500
00:57:36,330 --> 00:57:45,740
First is, why bundle conductors are used in
EHV lines. Second question is, what is
501
00:57:45,740 --> 00:57:54,060
transposition. And third question is, how
the effect of earth return current is taken
502
00:57:54,060 --> 00:57:58,000
into
account in inductance calculation for a three
503
00:57:58,000 --> 00:58:09,010
phase line with ground return system.
Especially, when the system is carrying, unbalanced
504
00:58:09,010 --> 00:58:17,320
current, so with this, we finish this
lesson.
505
00:58:17,320 --> 00:58:23,420
Thank you very much.
We will meet again for lesson 5. In which,
506
00:58:23,420 --> 00:58:48,050
we will talk about calculating the capacitance
of the transmission line.
507
00:58:48,050 --> 00:59:27,020
Thank you.
508
00:59:27,020 --> 00:59:27,020
.